Max Min Word Problems
Our approach to max min word problems is modeled after our
approach to related rates word problems. We will
Max Min Word Problems
Our approach to max min word problems is modeled after our
approach to related rates word problems. We will
1. draw a sketch of the situation;
Max Min Word Problems
Our approach to max min word problems is modeled after our
approach to related rates word problems. We will
1. draw a sketch of the situation;
2. label every quantity that can vary with a letter;
Max Min Word Problems
Our approach to max min word problems is modeled after our
approach to related rates word problems. We will
1. draw a sketch of the situation;
2. label every quantity that can vary with a letter;
3. write down the information of the problem in terms of those
letters;
Max Min Word Problems
Our approach to max min word problems is modeled after our
approach to related rates word problems. We will
1. draw a sketch of the situation;
2. label every quantity that can vary with a letter;
3. write down the information of the problem in terms of those
letters;
4. write down other relevant facts;
Max Min Word Problems
Our approach to max min word problems is modeled after our
approach to related rates word problems. We will
1. draw a sketch of the situation;
2. label every quantity that can vary with a letter;
3. write down the information of the problem in terms of those
letters;
4. write down other relevant facts;
5. restate and solve the problem;
Max Min Word Problems
Our approach to max min word problems is modeled after our
approach to related rates word problems. We will
1. draw a sketch of the situation;
2. label every quantity that can vary with a letter;
3. write down the information of the problem in terms of those
letters;
4. write down other relevant facts;
5. restate and solve the problem;
6. guarantee that our solution is as claimed.
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangular
garden. One side will be against a long barn. He has 100 ft of
fencing that he will use to surround the other three sides. What
are the dimensions of the garden of maximum area?
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangular
garden. One side will be against a long barn. He has 100 ft of
fencing that he will use to surround the other three sides. What
are the dimensions of the garden of maximum area?
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangular
garden. One side will be against a long barn. He has 100 ft of
fencing that he will use to surround the other three sides. What
are the dimensions of the garden of maximum area?
x + 2y = 100
y
x
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangular
garden. One side will be against a long barn. He has 100 ft of
fencing that he will use to surround the other three sides. What
are the dimensions of the garden of maximum area?
x + 2y = 100
Area = A = xy =
(100 − 2y )y =
100y − 2y 2 .
y
x
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangular
garden. One side will be against a long barn. He has 100 ft of
fencing that he will use to surround the other three sides. What
are the dimensions of the garden of maximum area?
x + 2y = 100
Area = A = xy =
(100 − 2y )y =
100y − 2y 2 .
We wish to
find the maximum value
of A and the values of x
and y that produce that
maximum value of A.
y
x
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangular
garden. One side will be against a long barn. He has 100 ft of
fencing that he will use to surround the other three sides. What
are the dimensions of the garden of maximum area?
x + 2y = 100
Area = A = xy =
(100 − 2y )y =
100y − 2y 2 .
We wish to
find the maximum value
of A and the values of x
and y that produce that
maximum value of A.
x ≥ 0 and y ≥ 0.
y
x
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangular
garden. One side will be against a long barn. He has 100 ft of
fencing that he will use to surround the other three sides. What
are the dimensions of the garden of maximum area?
x + 2y = 100
Area = A = xy =
(100 − 2y )y =
100y − 2y 2 .
We wish to
find the maximum value
of A and the values of x
and y that produce that
maximum value of A.
y
x
x ≥ 0 and y ≥ 0. Since x = 100 − 2y , we see that 100 − 2y ≥ 0.
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangular
garden. One side will be against a long barn. He has 100 ft of
fencing that he will use to surround the other three sides. What
are the dimensions of the garden of maximum area?
x + 2y = 100
Area = A = xy =
(100 − 2y )y =
100y − 2y 2 .
We wish to
find the maximum value
of A and the values of x
and y that produce that
maximum value of A.
y
x
x ≥ 0 and y ≥ 0. Since x = 100 − 2y , we see that 100 − 2y ≥ 0.
So 100 ≥ 2y and 50 ≥ y .
Biggest Garden
Example 1. Problem. A farmer plans to make a rectangular
garden. One side will be against a long barn. He has 100 ft of
fencing that he will use to surround the other three sides. What
are the dimensions of the garden of maximum area?
x + 2y = 100
Area = A = xy =
(100 − 2y )y =
100y − 2y 2 .
We wish to
find the maximum value
of A and the values of x
and y that produce that
maximum value of A.
y
x
x ≥ 0 and y ≥ 0. Since x = 100 − 2y , we see that 100 − 2y ≥ 0.
So 100 ≥ 2y and 50 ≥ y . That is, 0 ≤ y ≤ 50.
Biggest Garden
So we use our techniques for finding the maximum value of a
function on a closed interval when the function is continuous there.
Biggest Garden
So we use our techniques for finding the maximum value of a
function on a closed interval when the function is continuous there.
These apply since 100y − 2y 2 is a polynomial and so continuous
everywhere.
Biggest Garden
So we use our techniques for finding the maximum value of a
function on a closed interval when the function is continuous there.
These apply since 100y − 2y 2 is a polynomial and so continuous
everywhere.
dA
= 100 − 4y .
dy
Biggest Garden
So we use our techniques for finding the maximum value of a
function on a closed interval when the function is continuous there.
These apply since 100y − 2y 2 is a polynomial and so continuous
everywhere.
dA
= 100 − 4y .
dy
dA
= 0 iff y = 25.
A is differentiable everywhere, and
dy
Biggest Garden
So we use our techniques for finding the maximum value of a
function on a closed interval when the function is continuous there.
These apply since 100y − 2y 2 is a polynomial and so continuous
everywhere.
dA
= 100 − 4y .
dy
dA
= 0 iff y = 25.
A is differentiable everywhere, and
dy
If y = 0, then A = 0.
Biggest Garden
So we use our techniques for finding the maximum value of a
function on a closed interval when the function is continuous there.
These apply since 100y − 2y 2 is a polynomial and so continuous
everywhere.
dA
= 100 − 4y .
dy
dA
= 0 iff y = 25.
A is differentiable everywhere, and
dy
If y = 0, then A = 0.
If y = 25, then A = 100 · 25 − 2 · 252 = (100 − 50)25 > 0.
Biggest Garden
So we use our techniques for finding the maximum value of a
function on a closed interval when the function is continuous there.
These apply since 100y − 2y 2 is a polynomial and so continuous
everywhere.
dA
= 100 − 4y .
dy
dA
= 0 iff y = 25.
A is differentiable everywhere, and
dy
If y = 0, then A = 0.
If y = 25, then A = 100 · 25 − 2 · 252 = (100 − 50)25 > 0.
If y = 50, then x = 100 − 2 · 50 = 0, and A = 0 · 50 = 0.
Biggest Garden
So we use our techniques for finding the maximum value of a
function on a closed interval when the function is continuous there.
These apply since 100y − 2y 2 is a polynomial and so continuous
everywhere.
dA
= 100 − 4y .
dy
dA
= 0 iff y = 25.
A is differentiable everywhere, and
dy
If y = 0, then A = 0.
If y = 25, then A = 100 · 25 − 2 · 252 = (100 − 50)25 > 0.
If y = 50, then x = 100 − 2 · 50 = 0, and A = 0 · 50 = 0.
So the maximum area occurs with y = 25 and x = 100 − 2y = 50.
Closest Point
Example 2. Problem. Find the point on y = x 2 closest to the
point h3, 0i.
Closest Point
Example 2. Problem. Find the point on y = x 2 closest to the
point h3, 0i.
We let hx, y i be a point on the curve, and we let
d be the distance between hx, y i and h3, 0i.
6
5
4
y
3
2
d
1
K3
K2
K1
0
K1
1
x
2
3
Closest Point
Example 2. Problem. Find the point on y = x 2 closest to the
point h3, 0i.
We let hx, y i be a point on the curve, and we let
d bep
the distance between hx,p
y i and h3, 0i. So
2
2
d = (x − 3) + (y − 0) = (x − 3)2 + y 2 .
6
5
4
y
3
2
d
1
K3
K2
K1
0
K1
1
x
2
3
Closest Point
Example 2. Problem. Find the point on y = x 2 closest to the
point h3, 0i.
We let hx, y i be a point on the curve, and we let
d bep
the distance between hx,p
y i and h3, 0i. So
2
2
d = (x − 3) + (y − 0) = (x − 3)2 + y 2 .
We want to find the point hx, y i
that gives the minimum value for the distance d.
6
5
4
y
3
2
d
1
K3
K2
K1
0
K1
1
x
2
3
Closest Point
Example 2. Problem. Find the point on y = x 2 closest to the
point h3, 0i.
We let hx, y i be a point on the curve, and we let
d bep
the distance between hx,p
y i and h3, 0i. So
2
2
d = (x − 3) + (y − 0) = (x − 3)2 + y 2 .
We want to find the point hx, y i
that gives the minimum value for the distance d.
6
5
4
y
3
Remark
2
We let s = d 2 . So s = (x − 3)2 + y 2 .
1
K3
K2
K1
0
K1
d
1
x
2
3
Closest Point
Example 2. Problem. Find the point on y = x 2 closest to the
point h3, 0i.
We let hx, y i be a point on the curve, and we let
d bep
the distance between hx,p
y i and h3, 0i. So
2
2
d = (x − 3) + (y − 0) = (x − 3)2 + y 2 .
We want to find the point hx, y i
that gives the minimum value for the distance d.
6
5
4
y
3
Remark
2
We let s = d 2 . So s = (x − 3)2 + y 2 . The
minimum value for s and the minimum value for
d will occur at the same point. (Why?)
1
K3
K2
K1
0
K1
d
1
x
2
3
Closest Point
Example 2. Problem. Find the point on y = x 2 closest to the
point h3, 0i.
We let hx, y i be a point on the curve, and we let
d bep
the distance between hx,p
y i and h3, 0i. So
2
2
d = (x − 3) + (y − 0) = (x − 3)2 + y 2 .
We want to find the point hx, y i
that gives the minimum value for the distance d.
6
5
4
y
3
Remark
2
We let s = d 2 . So s = (x − 3)2 + y 2 . The
minimum value for s and the minimum value for
d will occur at the same point. (Why?) So we
will work with s instead of d.
1
K3
K2
K1
0
K1
d
1
x
2
3
Closest Point
Example 2. Problem. Find the point on y = x 2 closest to the
point h3, 0i.
We let hx, y i be a point on the curve, and we let
d bep
the distance between hx,p
y i and h3, 0i. So
2
2
d = (x − 3) + (y − 0) = (x − 3)2 + y 2 .
We want to find the point hx, y i
that gives the minimum value for the distance d.
6
5
4
y
3
Remark
2
We let s = d 2 . So s = (x − 3)2 + y 2 . The
minimum value for s and the minimum value for
d will occur at the same point. (Why?) So we
will work with s instead of d. This approach is
frequently useful in max min distance problems.
1
K3
K2
K1
0
K1
d
1
x
2
3
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Note that s is a polynomial and so continuous everywhere.
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Note that s is a polynomial and so continuous everywhere.
As usual, we compute the derivative of s.
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Note that s is a polynomial and so continuous everywhere.
As usual, we compute the derivative of s.
So
ds
= 2(x − 3) + 4x 3 =
dx
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Note that s is a polynomial and so continuous everywhere.
As usual, we compute the derivative of s.
ds
= 2(x − 3) + 4x 3 = 2(2x 3 + x − 3) =
dx
2(x − 1)(2x 2 + 2x + 3) =
So
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Note that s is a polynomial and so continuous everywhere.
As usual, we compute the derivative of s.
ds
= 2(x − 3) + 4x 3 = 2(2x 3 + x − 3) =
dx
2(x − 1)(2x 2 + 2x + 3) = 4(x − 1)(x 2 + x + 23 ) =
So
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Note that s is a polynomial and so continuous everywhere.
As usual, we compute the derivative of s.
ds
= 2(x − 3) + 4x 3 = 2(2x 3 + x − 3) =
dx
2(x − 1)(2x 2 + 2x + 3) = 4(x − 1)(x 2 + x + 23 ) =
4(x − 1)(x 2 + 2( 21 )x + 14 + 54 ) =
So
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Note that s is a polynomial and so continuous everywhere.
As usual, we compute the derivative of s.
ds
= 2(x − 3) + 4x 3 = 2(2x 3 + x − 3) =
dx
2(x − 1)(2x 2 + 2x + 3) = 4(x − 1)(x 2 + x + 23 ) =
4(x − 1)(x 2 + 2( 21 )x + 14 + 54 ) = 4(x − 1)((x + 12 )2 + 54 ).
So
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Note that s is a polynomial and so continuous everywhere.
As usual, we compute the derivative of s.
ds
= 2(x − 3) + 4x 3 = 2(2x 3 + x − 3) =
dx
2(x − 1)(2x 2 + 2x + 3) = 4(x − 1)(x 2 + x + 23 ) =
4(x − 1)(x 2 + 2( 21 )x + 14 + 54 ) = 4(x − 1)((x + 12 )2 + 54 ).
So
Remark
Let g (x) = 2x 3 + x − 3.
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Note that s is a polynomial and so continuous everywhere.
As usual, we compute the derivative of s.
ds
= 2(x − 3) + 4x 3 = 2(2x 3 + x − 3) =
dx
2(x − 1)(2x 2 + 2x + 3) = 4(x − 1)(x 2 + x + 23 ) =
4(x − 1)(x 2 + 2( 21 )x + 14 + 54 ) = 4(x − 1)((x + 12 )2 + 54 ).
So
Remark
Let g (x) = 2x 3 + x − 3. Note that g (1) = 0.
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Note that s is a polynomial and so continuous everywhere.
As usual, we compute the derivative of s.
ds
= 2(x − 3) + 4x 3 = 2(2x 3 + x − 3) =
dx
2(x − 1)(2x 2 + 2x + 3) = 4(x − 1)(x 2 + x + 23 ) =
4(x − 1)(x 2 + 2( 21 )x + 14 + 54 ) = 4(x − 1)((x + 12 )2 + 54 ).
So
Remark
Let g (x) = 2x 3 + x − 3. Note that g (1) = 0. So (x − 1) divides
2x 3 + x − 3.
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Note that s is a polynomial and so continuous everywhere.
As usual, we compute the derivative of s.
ds
= 2(x − 3) + 4x 3 = 2(2x 3 + x − 3) =
dx
2(x − 1)(2x 2 + 2x + 3) = 4(x − 1)(x 2 + x + 23 ) =
4(x − 1)(x 2 + 2( 21 )x + 14 + 54 ) = 4(x − 1)((x + 12 )2 + 54 ).
So
Remark
Let g (x) = 2x 3 + x − 3. Note that g (1) = 0. So (x − 1) divides
2x 3 + x − 3. We find the quotient by long division.
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Note that s is a polynomial and so continuous everywhere.
As usual, we compute the derivative of s.
ds
= 2(x − 3) + 4x 3 = 2(2x 3 + x − 3) =
dx
2(x − 1)(2x 2 + 2x + 3) = 4(x − 1)(x 2 + x + 23 ) =
4(x − 1)(x 2 + 2( 21 )x + 14 + 54 ) = 4(x − 1)((x + 12 )2 + 54 ).
So
Remark
Let g (x) = 2x 3 + x − 3. Note that g (1) = 0. So (x − 1) divides
2x 3 + x − 3. We find the quotient by long division. Also, we find
that x 2 + x + 23 = (x + 21 )2 + 54 by “completing the square”.
Closest Point
Now s = (x − 3)2 + y 2 = (x − 3)2 + x 4 since y = x 2 .
Note that s is a polynomial and so continuous everywhere.
As usual, we compute the derivative of s.
ds
= 2(x − 3) + 4x 3 = 2(2x 3 + x − 3) =
dx
2(x − 1)(2x 2 + 2x + 3) = 4(x − 1)(x 2 + x + 23 ) =
4(x − 1)(x 2 + 2( 21 )x + 14 + 54 ) = 4(x − 1)((x + 12 )2 + 54 ).
So
Remark
Let g (x) = 2x 3 + x − 3. Note that g (1) = 0. So (x − 1) divides
2x 3 + x − 3. We find the quotient by long division. Also, we find
that x 2 + x + 23 = (x + 21 )2 + 54 by “completing the square”.
Closest Point
(x + 12 )2 +
5
4
≥
5
4
> 0 for all x.
Closest Point
(x + 12 )2 + 54 ≥ 54 > 0 for all x.
So if x < 1, then x − 1 < 0,
Closest Point
(x + 12 )2 + 54 ≥ 54 > 0 for all x.
So if x < 1, then x − 1 < 0, and so
ds
= 4(x − 1)((x + 21 )2 + 54 ) < 0.
dx
Closest Point
(x + 12 )2 + 54 ≥ 54 > 0 for all x.
So if x < 1, then x − 1 < 0, and so
ds
= 4(x − 1)((x + 21 )2 + 54 ) < 0.
dx
So s is decreasing on (−∞, 1].
Closest Point
(x + 12 )2 + 54 ≥ 54 > 0 for all x.
So if x < 1, then x − 1 < 0, and so
ds
= 4(x − 1)((x + 21 )2 + 54 ) < 0.
dx
So s is decreasing on (−∞, 1].
So if x > 1, then x − 1 > 0,
Closest Point
(x + 12 )2 + 54 ≥ 54 > 0 for all x.
So if x < 1, then x − 1 < 0, and so
ds
= 4(x − 1)((x + 21 )2 + 54 ) < 0.
dx
So s is decreasing on (−∞, 1].
So if x > 1, then x − 1 > 0, and so
ds
= 4(x − 1)((x + 21 )2 + 54 ) > 0.
dx
Closest Point
(x + 12 )2 + 54 ≥ 54 > 0 for all x.
So if x < 1, then x − 1 < 0, and so
ds
= 4(x − 1)((x + 21 )2 + 54 ) < 0.
dx
So s is decreasing on (−∞, 1].
So if x > 1, then x − 1 > 0, and so
ds
= 4(x − 1)((x + 21 )2 + 54 ) > 0.
dx
So s is increasing on [1, ∞).
Closest Point
(x + 12 )2 + 54 ≥ 54 > 0 for all x.
So if x < 1, then x − 1 < 0, and so
ds
= 4(x − 1)((x + 21 )2 + 54 ) < 0.
dx
So s is decreasing on (−∞, 1].
So if x > 1, then x − 1 > 0, and so
ds
= 4(x − 1)((x + 21 )2 + 54 ) > 0.
dx
So s is increasing on [1, ∞).
So the absolute minimum value for s (and d) occurs with x = 1.
Closest Point
(x + 12 )2 + 54 ≥ 54 > 0 for all x.
So if x < 1, then x − 1 < 0, and so
ds
= 4(x − 1)((x + 21 )2 + 54 ) < 0.
dx
So s is decreasing on (−∞, 1].
So if x > 1, then x − 1 > 0, and so
ds
= 4(x − 1)((x + 21 )2 + 54 ) > 0.
dx
So s is increasing on [1, ∞).
So the absolute minimum value for s (and d) occurs with x = 1.
So h1, 1i is the point on y = x 2 closest to h3, 0i.